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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A uniform ladder of mass 10 kg leans against a smooth vertical wall making an angle 53^(@) with it. The other end rests on a rough horizontal floor. Find the normal force and the frictional force that the floor exerts on the ladder. |
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Answer» Solution :The LADDER is in EQUILIBRIUM. `THEREFORE N_(1)=f` and `N_(2)=W` Taking torque about .B. `N_(1)(AO)=W(CB)`. or `N_(1)(AB) cos 53 = W((AB)/(2))sin 53` or `N_(1)=(2)/(3)W` and `N_(2)=W=10xx9.8=98 N`. The frictional force is `f=N_(1)=(2)/(3)W=65 N` 
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| 2. |
Why do spring balances show wrongreading after long use ? |
| Answer» SOLUTION :After long use the spring balance gets fatigue and losss its elastic strength . ASA result the extension of the spring for a GIVEN load is more and hence it GIVES wrong readings . | |
| 3. |
During melting process, the heat given to a solid is used in (generally) |
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Answer» Increasing the temperature |
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| 4. |
A flywheelof mass 25kg has a radius of 0.2 m. What force should be applied to the rim so that it acquires an angular acceleration of5rads^(-2)? |
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Answer» Solution :The MI of the FLYWHEEL= 1/2`Mr^2` =1/2`TIMES` 25 `times(0.2)^2` = 1/2 TORQUE `tau`= `Ialpha`=1/2 `times`5 |
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| 5. |
The displacement of a particle executing SHM is given by Y = 10 sin (3t + pi//3 )m and 't' is in seconds. The initial displacement and maximum velocity of the particle are respectively |
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Answer» ` 5sqrt(3)` m and `30m//"SEC"` |
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| 6. |
A smooth sphere is moving on a horizontal surface with velocity vector 2hati+2hatj immediately before it hits a vertical wall. The wall is parallel to hatj vector and the coefficient of restitution between the sphere and the wall is e = 1//2. The velocity vector of the sphere after it hits the wall is |
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Answer» `HATI-HATJ` |
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| 7. |
In the above question, the reasonfor the bodiesto havedifferenttimes of descent is |
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Answer» THEYHAVE same MASS |
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| 8. |
An athlete running on a straight track is an example for the whirling motion of a stone attached to a string is a............... motion. |
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Answer» linear |
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| 9. |
Three bodies A , B and C of masses, m, m and sqrt3 respectively are supplied heatat a constant rate, The change in temperature thetaversus time t graph for A,B and C are shown by I , II respectively.If their specific heat capacities are S_A , S_(B) and S_Crespectively . then which of the following relation is correct ? (initial temperature of the body is 0^0C ). |
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Answer» `S_(A) GT S_(B) gt S_C` |
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| 10. |
IF theerrorin themeasurementofmassand speedof aparticleare |
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Answer» SOLUTION :kineticenergy`K = (1)/(2) mv ^(2) ` ` THEREFORE `Fractionalerror ,`( DELTAK ) /( K )= ( Deltam ) /(m )+ 2 (Delta v ) /(v) ` `= 0.1%+ 2xx0.2%= 0.5% ` |
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| 11. |
Universal gravitational constant, universal gas constant and …….. Are some of the ………… constant. |
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Answer» |
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| 12. |
A tunnel is dug along the diameter of the earth. The gravitational force on particle of mass m placed in the tunnel at a distance .x. from the centre . |
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Answer» Solution :MASS of EARTH `M=4/3 piR^(3)rho` Mass of imaginary sphere, having radiu x, `M^(1)=4/3pix^3rho("or")M^(1)/M=x^3/R^3` `:.` GRAVITATIONAL force on .m. DUE to `M_1` `F=(GM^1m)/x^2("or") F=(Gm)/x^2(x^3/R^3M)=(GMmx)/R^3` 
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| 13. |
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to done against the gravitational force between them to take the particle is away from the sphere. |
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Answer» Solution :`U = (-GMm)/R = (-6.67 xx10^(-11) XX 100 xx 10 xx 10^(-3))/(10 xx 10^(-2))` `U = 6.67 xx10^(-10)J ` So ,the MAOUNT of workdone totakethe particleuptoinfinitewill be` 6 . 67 xx10^(-10)` J |
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| 14. |
A copper wire of negligible mass, length (l), corss-sectionalarea (A) is kept on a smooth horizontal table with one end fixed, a ball of mass .m. is attached at other end. The wire and the ball are rotated with angular velocity .omega.. If wire elongates by Deltal then find Young.s modulus of wire. If on increasing the angular velocity from omega to omega^(1) the wire breakdown, obtain breaking stress (Deltal lt lt l) |
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Answer» Solution :(a) `r=l+Deltal` `F=T=mrw^(2)=m(l+Deltal)omega^(2)` as `Deltal` in SMALL, `F=mlomega^(2)""y=(Fl)/(AE)""y=((mlomega^(2))l)/(A Deltal)` (b) We know Breakingstress `=("Breaking FORCE")/("Area of cross section")=(ml omega^(1^(2)))/(A)` |
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| 15. |
Find the kinetic energy needed to project a body of mass m from the centre of a ring of mass M and radius R so that it will never comne back. |
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Answer» |
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| 16. |
Two sides of a triangle are given by hati+hatj+hatk and -hati+2hatj+3hatk then area of triangle is |
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Answer» `SQRT(26)` |
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| 17. |
A celsius and Fahrenheit thermometer are put in a hot bath. The reading on Fahrenheit thermometer is just three times the reading on Celsius thermometer . What is the temperature of the bath? |
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Answer» |
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| 18. |
A metal wire is bent in the form of a ring with a small gap at the free ends. If the wire is heated what happens to the gap? |
| Answer» SOLUTION :INCREASES | |
| 19. |
A clock regulated by a seconds pendulum keeps correct time. During summer the length of the pendulum increases to 1.02m. How much will the clock gain or lose in one day ? |
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Answer» 2280 SEC |
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| 20. |
Discuss elastic collision in two dimesnsion . |
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Answer» Solution :As shownin figure , supposea ball of mass `m_(1)` moving in X - direction with SPEED `v_(1i)` collideselastically with a STATIONARY ballof mass `m_(2)` . after the collision theseballmovein the directionmakingangle `theta_(1) and theta_(2)` with the X - axis with velocities `v_(1f) and v_(2F)` Momentum is conserved in collision , ` :. ` Momentum beforecollision= momentum after collision ` :. m_(1)v_(1i) = m_(1)v_(1f) +m_(2)v_(2f) ""...(1) [ :. v_(2i)=0 ] ` Here , there are three equations (2) , (3) and (4) and from theseunknown quantities can bedetermined . usually the value of `m_(1),m_(2),v_(1i) and v_(2i)` are known , whereas four terms `v_(1f),v_(2f) and theta_(1) and theta_(2)` are quantitiesmust be knownas three equations can be givethe valuesof only three unknownquantities . If the motion in one dimension then unknown terms `v_(1f) and v_(2f)` are only be obtainedbecause in this motion `theta_(1)=theta_(2) = 0 ` and henceunknown TERM can be knownby the equations in one dimension . |
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| 21. |
The phase of simple harmonic motion at t=0is called |
| Answer» Answer :D | |
| 22. |
An ideal gas undergoes a process in which PV^(-a)= constant, where V is the volume occupied by the gas initially at pressure P. At the end of the process, rms speed of gas molecules has become a^(1//2)times of its initial value. What will be the value of C_vso that energy transferred by the heat to the gas is .a. times of the initial energy. (Delta Q = aPV) |
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Answer» `((a^2 + 1)R)/(a^2 - 1)` |
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| 23. |
Centripetal acceleration is |
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Answer» a constant VECTOR |
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| 24. |
The system is in equilibrium if the spring balance is calibrated in newton what does it record in each case |
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Answer» Solution :(a)oneweightacts as supportanotheractsas weight Sotension T=10g=10`XX 10` T= 100N (B)`T= 2xx 10xx g` `=2 xx 10xx 10 = 200N` (c )`T = 10 xx 10 sin 30^(@)` `=10 xx 10 xx (1)/(2) = 50N` |
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| 25. |
Any rigid body can have more than one value of moment of inertia? |
| Answer» SOLUTION :YES. Moment of inertia is different for the chosen different AXIS. | |
| 26. |
A particle encute SHM between x = -A and x = +A, The time taken for it to go from 0 to A//2 is T_(2) and to go from A//2 to A is T_(2), then |
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Answer» `T_(1) lt T_(2)` |
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| 27. |
One end of a uniform rod of length 1 m is placed in boiling water while its other end is placed in melting ice. A point P on the rod is maintained at a constant temperature of 800^(0)C. The mass of steam produced per seecond is equal to the mass of ice melted per second. If specific latent heat of steam is 7 times the specific latent heat of ice. Then the distance .p. form the steam chamber is (1)/(K). The value of K = ______________ |
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| 28. |
A solid sphere and a hallow sphere of same mass and radius rotate about their diameter with same K.E. if torques of eaual magnitude are applied on them, the ratio angular displacement before they are brought to rest is |
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Answer» `1:1` |
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| 29. |
The time period of a particle performing linear SHM is 12 s. What is the time taken by it to make a displacement equal to half its amplitude? |
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Answer» 1 sec |
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| 30. |
A soap bubble of radius r and surface tension constant T is given a charge, so that its surface charge density is sigma. Due to charge, the radius os the soap bubble becomes double then find 'sigma'. (atmospheric pressure = P_(0)) |
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Answer» Solution :initial pressure inside the BUBBLE `P_(i) = P_(0) + (4T)/(r)` Now a uniform surface charge in given to the bubble The surface tension is a pulling a force, which increases pressure inside the bubble (by `(4T)/(r))` but the charge given to the surface will repel each other. So due to the charge given. pressure inside the bubble will decrease (by `(SIGMA^(2))/(2epsilon_(0))`) So, final pressure inside the bubble `P_(1) = P_(0) + (4T)/(r_(f)) - (sigma^(2))/(2epsilon)` As the TEMPERATURE of the gas inside the bubble if costant so, `P_(1)V_(1) = P_(1)V_(1)` `(P_(0) + (2T)/(r))((4)/(3)pir^(3)) = (P_(0) + (4T)/(r_(f))) = (P_(0) + (4T)/(r_(f)) - (sigma^(2))/(2epsilon_(0)))((4)/(3)pir_(f)^(3))` Here Put`r_(1) = 2r` So, get `sigma = sqrt((7P_(0) - (12T)/(r))2epsilon)`. |
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| 31. |
The driver of a car moving towards a rocket launching with a speed of 6ms^(-1) observed that the rocket is moving with speed of 10 ms^(-1). The upward speed of the rocket as seen by the stationary observer is nearly |
| Answer» Answer :C | |
| 32. |
Statement-1:Susceptibility is expressed as Am^(-1) Statement-2:Magnetic flux is expressed as JA^(-1). |
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Answer» Statement-1 is true and statement -2 is true |
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| 33. |
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^(-1). Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^(-1). What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !] |
| Answer» SOLUTION :(a) `5KM h^(-1), 5km h^(-1)`,(b) `0.6 km h^(-1)`, (c ) `(15)/(8)km h^(-1), (45)/(8) km h^(-1)` | |
| 34. |
All surface are smooth. Find the acceleration of mass m relative to wedge when wedge is moving with acceleration 'a' |
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Answer» Solution :Net force acting down the plane is `F=ma cos THETA + MG SIN theta` ACCELERATION down the plane `a=g sin theta + a cos theta` |
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| 35. |
The acceleration of a particle is given by vec(a) = [2 hat(i) + 6t hat(j) + (2pi^(2))/(9) cos ((pi t)/(3)) hat(k)] ms^(-2). The initial conditions are vec(r ) (0) = x(0)hat(i) + y(0) hat(j) + z(0) hat(k) = vec(0), vec(v) (0) = (2 hat(i) + hat(j)) ms^(-1) wher vec(v) (0) = (d hat(i) (0))/(dt) = (dx (0))/(dt) hat(i) + (dy(0))/(dt) hat(j) + (dz(0))/(dt) hat(k) The position vector at t= 2 s is |
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Answer» `(3 HAT(i) + 8hat(j) + 10 hat(k))` |
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| 36. |
The force acting tangential to the layers of the liquid is called: |
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Answer» BUOYANT force |
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| 37. |
In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat? |
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Answer» EAST (APPROXIMATELY) |
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| 38. |
0.8 litre of water flows in 5 minutes in a capillary tube of length 10 cm and of diameter of 1.5 mm. The height of the water surface is 45 cm from the axis of the capillary tube. Find the coefficient of viscosity of water. |
| Answer» SOLUTION :`2.056xx10^(-3) P_a s` | |
| 39. |
A particle moves in a circular path with a uniform speed. Its motion is |
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Answer» periodic |
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| 40. |
A horizontal force applied on a body on a rough horizontal surface produces an acceleration 'a'. If coefficient of friction between the body & surface which is m is reduced to m/3, the accele-ration increases by 2 units. The value of m is |
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Answer» `2//3g` |
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| 41. |
Statements : (a) Position of C.m depends on acceleration due to gravity where as position of C.g independent on acceleration due to gravity. (b) Centre of gravity of a body is defined to know the amount of stability of the body and the center of mass of the body is defined to describe the nature of motion of the body. |
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Answer» a is TRUE, B is FALSE |
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| 42. |
For a projectile .R. is range and .H. is maximum height |
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Answer» a-g, b-h, C-e, d-f |
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| 43. |
A player caught a cricket ball of mass 150g moving at a rate of 20 m/s. If the catching process is completed in 0.1s, the force of the blow exerted by the ball on the hand of the player is equal to : |
| Answer» ANSWER :C | |
| 44. |
A thermometer is taken from the melting ice into a warm liquid which results in rise in mercury level to 1/5thof the distance between the lower and the upper fixed points. Calculate the temperature of liquid in kelvins. |
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Answer» Solution : If the distance between the lower and upper FIXED POINTS is 1 UNIT, then Rise in mercury level =`1/5`UNITS `(1/5 - 0)/(1 - 0) = (T_C - 0)/(100)` ` rArr T_C = 20^@C` or`T = 20 + 273.15 = 293.15K` |
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| 45. |
A hollow cylinder of inner radius 4 cm and outer radius is 5cm and a solid cylinder of radius 3 cm are under the same load. They are of the same material and of same length, Find the ratio of their clongations. |
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Answer» |
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| 46. |
A particle of mass 0.2 kg is moving with linear velocity (i-j+2k). If the radius vector r=4i+j-k, the angular momentum of the particle is |
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Answer» `2.14` UNITS |
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| 47. |
In the formula, N=-D[(n_(2)-n_(1))/(x_(2)-x_(1))], D - Diffusion coefficient, n_(1) and_(2) is number of molecules in unit volume along x_(1) and x_(2) which represents distances where N is number of molecules passing through per unit area per unit time. Calculate dimensions of D. |
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Answer» Solution :By HOMOGENEITY theory of dimensions, `"Dimensions of (N) = dimension of D "XX("dimension of "(n_(2)-n_(1)))/("dimension of "(x_(2)-x_(1)))` `(1)/(L^(2)T)="dimensions of D"xx(L^(-3))/(L)` `"Dimensions of D"=(L)/(L^(-3)XXL^(2)T)=(L^(2))/(T)` `=[M^(0)L^(2)T^(-1)]` |
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| 48. |
A copper block of mass 1 Kg slides down on a rough inclined plane of inclination 37^(0) at a constant speed. Find the increase in temperature of the block as it slides down through 60 cm assuming that the loss in mechanical energy goes to the copper block as thermal energy? (specific heat of copper = 420 J Kg^(-1)K^(-1) and g = 10ms^(-2)) |
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Answer» Solution :Loss in gravitational PE = Thermal ENERGY: MG sin `370 = ms Deltatheta` `implies 10 XX 3/5 = 420 xx Deltatheta` , Change in temperature `Deltatheta = 8.6 xx 10^(-30)C` |
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| 49. |
A body of mass m is suspended by two strings making angles alpha_(1) and alpha_(2) with the horizontal. Find the tension in the strings. |
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Answer» Solution :As the weights are suspended from the midpoints, the tensions in the two sides are the same. If it is T then resultant of upward components of T = T sin`ALPHA + T sin alpha = 2 T " " sin alpha`. At equilibrium, the resultant has to BALANCE the weight. In this CASE, ` T_(1)= (W)/(2 sin alpha_(1)) and T_(2) = (W)/(2 sin alpha_(2))` As `alpha_(1) gt alpha_(2) , T_(2) gt T_(1),` i.e., the second string has a higher tension.  |
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